0
$\begingroup$

Let $x_{n+1}$=$\frac{\alpha x_n^2+\beta x_n(1-x_n)}{\alpha x_n^2+2\beta x_n(1-x_n)+\gamma(1-x_n)^2}$ be a sequence, where $0\leq \alpha,\beta,\gamma\leq 1$. $0

Suppose $\alpha>\beta>\gamma$ then show that $$ converge to 1.

My approach. I showed $0

The second part of the questions says, if $\alpha <\beta<\gamma$ then $$ converges to zero. I am guessing it will follow the same pattern as above.

  • 0
    Just note, I think the place you reach will not lead you to $x_{n+1} \geq x_n$ since $[x_n^2+\frac{\beta}{\alpha} x_n(1-x_n) ] - x_n = \frac{\beta}{\alpha} x_n(1-x_n) - x_n( 1-x_n) = x_n(1-x_n)( \frac{\beta}{\alpha} - 1) < 0 $ since $ \alpha > \beta $ so $ \frac{\beta}{\alpha} < 1 $. so $x_n^2+\frac{\beta}{\alpha} x_n(1-x_n) < x_n $ ! – 2017-01-18
  • 0
    Any idea on how to proceed or what to do next? – 2017-01-18

1 Answers 1

1

In order to show that $$ converges to $1$, enough to show that it is increasing, since it is bounded above by $1$. So , required to show $$ x_{n+1} \geq x_n$$ We have: $$x_{n+1}= x_n\Big[ \frac{\alpha x_n+\beta (1-x_n)}{\alpha x_n^2+2\beta x_n(1-x_n)+\gamma(1-x_n)^2} \Big]$$ So it is enough to show that $$ \frac{\alpha x_n+\beta (1-x_n)}{\alpha x_n^2+2\beta x_n(1-x_n)+\gamma(1-x_n)^2} \geq 1 $$ i.e. $$ \alpha x_n+\beta (1-x_n) \geq \alpha x_n^2+2\beta x_n(1-x_n)+\gamma(1-x_n)^2 $$ i.e. $$ \Big[\alpha x_n+\beta (1-x_n)\Big] - \Big[\alpha x_n^2+2\beta x_n(1-x_n)+\gamma(1-x_n)^2 \Big] \geq 0 $$ Indeed

\begin{align} \Big[\alpha x_n+\beta (1-x_n)\Big] - \Big[\alpha x_n^2+2\beta x_n(1-x_n)+\gamma(1-x_n)^2 \Big] \\ =\alpha x_n+\beta (1-x_n) - \alpha x_n^2-2\beta x_n(1-x_n)-\gamma(1-x_n)^2 \\ =\alpha x_n( 1-x_n) + \beta(1-x_n)(1-2x_n)-\gamma(1-x_n)^2 \\ =(1-x_n)(\alpha x_n+\beta -2\beta x_n -\gamma (1-x_n) ) \\ = (1-x_n) \Big( (\alpha- \beta) x_n + (\beta -\gamma)- (\beta -\gamma)x_n \Big) \\ =(1-x_n)\Big( (\alpha- \beta) x_n+(\beta -\gamma)(1- x_n ) \Big) \geq 0 \end{align} As required. I hope this helps you.